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Baltic Way 1995 · Problem 13

Combinatorics

Consider the following two person game. A number of pebbles are situated on the table. Two players make their moves alternately. A move consists of taking off the table xx pebbles where xx is the square of any positive integer. The player who is unable to make a move loses. Prove that there are infinitely many initial situations in which the second player can win no matter how his opponent plays.

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Topics

Colorings and configurations · Pigeonhole and extremal arguments · Games and strategies

Solutions

Solution

Solution:

Suppose that there is an nn such that the first player always wins if there are initially more than nn pebbles. Consider the initial situation with n2+n+1n^{2}+n+1 pebbles. Since (n+1)2>n2+n+1(n+1)^{2}>n^{2}+n+1, the first player can take at most n2n^{2} pebbles, leaving at least n+1n+1 pebbles on the table. By the assumption, the second player now wins. This contradiction proves that there are infinitely many situations in which the second player wins no matter how the first player plays.

Contest context

Results from Baltic Way 1995

9 teams

Mean score
1.7 / 5
Scores of 4 or 5
3 / 9
Estonia
0 / 5

Score distribution

06
10
20
30
40
53
All team scores
TeamScore
Poland5 / 5
Latvia5 / 5
Sweden0 / 5
Lithuania0 / 5
Denmark0 / 5
Finland5 / 5
St. Petersburg0 / 5
Estonia0 / 5
Iceland0 / 5